a space X is Hausdorff if and only if Δ⁢(X) is closed

Theorem.

Proof.

First, some preliminaries: Recall that the diagonal mapΔ:X→X×X is defined as x⟼Δ(x,x). Also recall that in a topology generated by a basis (like the product topology), a set Y is open if and only if, for every point y∈Y, there’s a basis element B with y∈B⊂Y. Basis elements for X×X have the form U×V where U,V are open sets in X.

Now, suppose that X is Hausdorff. We’d like to show its image under Δ is closed. We can do that by showing that its complementΔ⁢(X)c is open. Δ⁢(X) consists of points with equal coordinates, so Δ⁢(X)c consists of points (x,y) with x and y distinct.

For any (x,y)∈Δ⁢(X)c, the Hausdorff condition gives us disjoint open U,V⊂X with x∈U,y∈V. Then U×V is a basis element containing (x,y). U and V have no points in common, so U×V contains nothing in the image of the diagonal map: U×V is contained in Δ⁢(X)c. So Δ⁢(X)c is open, making Δ⁢(X) closed.

Now let’s suppose Δ⁢(X) is closed. Then Δ⁢(X)c is open. Given any (x,y)∈Δ⁢(X)c, there’s a basis element U×V with (x,y)∈U×V⊂Δ⁢(X)c. U×V lying in Δ⁢(X)c implies that U and V are disjoint.

If we have x≠y in X, then (x,y) is in Δ⁢(X)c. The basis element containing (x,y) gives us open, disjoint U,V with x∈U,y∈V. X is Hausdorff, just like we wanted.
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